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1-3-x-2-x-2-4x-dx-

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Question Number 174336 by cortano1 last updated on 30/Jul/22
∫_1 ^3 ((∣x−2∣)/(x^2 −4x)) dx =?
∫31∣x−2∣x2−4xdx=?
Answered by Mathspace last updated on 30/Jul/22
I=−∫_1 ^2 ((x−2)/(x^2 −4x))dx+∫_2 ^3 ((x−2)/(x^2 −4x))dx f(x)=((x−2)/(x(x−4)))=(a/x)+(b/(x−4)) a=((−2)/(−4))=(1/2) and b=((4−2)/4)=(1/2) ⇒∫_1 ^2 f(x)dx=(1/2)∫_1 ^2 ((1/x)+(1/(x−4)))dx =(1/2)[ln∣x∣+ln∣x−4∣]_1 ^2 =(1/2){2ln2−ln3} ∫_2 ^3 ((x−2)/(x^2 −4x))dx=(1/2)∫_2 ^3 ((1/x)+(1/(x−4)))dx =(1/2)[ln∣x∣+ln∣x−4∣]_2 ^3 =(1/2){ln3−2ln2} ⇒ ∫_1 ^3 ((∣x−2∣)/(x^2 −4x))dx=−(1/2){2ln2−ln3} +(1/2){ln3−2ln2} =−ln2+(1/2)ln3+(1/2)ln3−ln2 =ln3−2ln2 .
I=−∫12x−2x2−4xdx+∫23x−2x2−4xdxf(x)=x−2x(x−4)=ax+bx−4a=−2−4=12andb=4−24=12⇒∫12f(x)dx=12∫12(1x+1x−4)dx=12[ln∣x∣+ln∣x−4∣]12=12{2ln2−ln3}∫23x−2x2−4xdx=12∫23(1x+1x−4)dx=12[ln∣x∣+ln∣x−4∣]23=12{ln3−2ln2}⇒∫13∣x−2∣x2−4xdx=−12{2ln2−ln3}+12{ln3−2ln2}=−ln2+12ln3+12ln3−ln2=ln3−2ln2.

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